6x^2+11xy+4y^2 factor-able?

6 • 4 = 24

Factors of 24 whose sum is 11?

1, 12; 2, 12; 3, 8; 4, 6

3 and 8

4x^2 + 3xy + 8xy + 4y^2 = x(4x + 3y) + 2y(4x + y) = (4x + y)(x + 2y)

1) Considering the given expression as a quadratic in 'x',
a = 6; b = 11y; and c = 4y^2

2) ==> Discriminanat D = b^2 - 4ac, D = 121y^2 - 96y^2 = 25y^2

3) Since this discriminant D is a perfect square, the given expression is factorizable.

4) To factorize, split the middle term 11 = 8 + 3, so that 8 x 3 = 24, which is the product of first and last coefficients.

5) Hence, 6x^2+11xy+4y^2 = 6x^2 + 8xy + 3xy + 4y^2

==> = 2x(3x + 4y) + y(3x + 4y) = (3x + 4y)(2x + y)

So, 6x^2+11xy+4y^2 = (3x + 4y)(2x + y)

so the first thing has x^2 and the last thing has y^2, so the middle term must have both x and y to even think about factoring. For example, if the problem had been 6x^2+11x+4y^2 you know it can't be factored in the usual way (although you could factor out an x from the first two terms alone).

so all the ingredients are there, since the middle term does have x and y.

so you know it will look like this:

(_x+_y)(_x+_y)

now you just need to fill in the blanks.

you have a 6 in front of the x^2, so the numbers for the _ in front of the x's must be factors of 6. And the numbers in front of the y's must be factors of 4.

trial and error leads you to realize you need to factor the 6 as 3 and 2, and the 4 as 4 and 1 to get:

(3x+4y)(2x+1y)

6x^2 + 11xy + 4y^2
= 6x^2 + 8xy + 3xy + 4y^2
= (6x^2 + 8xy) + (3xy + 4y^2)
= 2x(3x + 4y) + y(3x + 4y)
= (3x + 4y)(2x + y)

(2x + 1y)(3x + 4y)

Basically you have to get 6x^2 as the first term, so the possible ones are 6x, and x, or 3x and 2x. You have to get 4y^2 as the last term, so the other two numbers are either 4y, 1y, or 2y and 2y. The middle term has to add up to 11, so the only possible one that works is 8 and 3. By educated guess and check you will see that the above factor works.